A 5.00-g bullet moving with an initial speed of v i = 390 m/s is fired into and

Question

A 5.00-g bullet moving with an initial speed of vi = 390 m/s is fired into and passes through a 1.00-kg block as shown in the figure below. The block, initially at rest on a frictionless, horizontal surface, is connected to a spring with force constant890 N/m. The block moves d = 4.40 cm to the right after impact before being brought to rest by the spring.

(a) Find the speed at which the bullet emerges from the block.
m/s

(b) Find the amount of initial kinetic energy of the bullet that is converted into internal energy in bullet–block system during the collision.
J

Explanation / Answer

Initial Kinetic Energy = 1/2 * mv^2
Initial Kinetic Energy = 1/2 * 5/1000 * 390^2 J
Initial Kinetic Energy = 380.25 J

Spring Potential Energy = 1/2 * kx^2
Spring Potential Energy = 1/2 * 890 * (4.40/100)^2
Spring Potential Energy = 0.8615 J

Now This Spring Potential Energy is Equal to the Kinetic Energy of (Block + Bulle)

1/2 * m*V^2 = 1/2 * kx^2
1/2 * (1.0) * V^2 = 0.8615
V = 1.31 m/s

Using Momentum Conservation,
Initial Momentum = Final Momentum
(5/1000) * 390 = 5/1000 * v + 1 * 1.31
v = 128 m/s

Final Kinetic Energy of bullet = 1/2 * mv^2
Final Kinetic Energy of bullet = 1/2 * 5/1000 * 128^2 J
Final Kinetic Energy of bullet = 40.96 J

Amount of Initial Kinetic Energy converted into internal energy in bullet–block system during the collision
= 380.25 - 40.96
= 339.3 J

(a) Speed at which Bullet emerges the Block, v = 128 m/s.
(b) 339.3 J

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